pass a value to another page

nllewellyn

Hi,

I am trying to create a search feature my website. I worked on the query and an results page first and everything works OK, thanks to the help I received from this forum to fix the paging problem.

As soon and I introduced a form into the scenarion things have gone pear shaped.

I enter a search term and submit.
the search term is passed to my results page and the first 10 rows are returned along with a NEXT>> link.

So far so good 🙂

When I click on the next link, the next page returns no results. Even though I know there are 60 rows to return.

The query is:

 $query = "SELECT * FROM $dbtable where station like '$search' limit $offset, $limit";

It seem as though the search variable isnt being passed to the next page, So fails to return the next rows.

how can I pass the variable to the next page?

Thanks

yelvington

URL-encoded.

Or in cookies.

Or in session variables.

nllewellyn

Thanks for your suggestions

I am already passing some URL encoded values:

  if (!($num_results < $limit))
   {
    // not last page so give NEXT link
    $newoffset=$offset+$limit;
    echo "<a href=\"$PHP_SELF?offset=$newoffset\">NEXT</a><p>\n";
	}

but I'm not sure how to add the extra values. The code above generates my next page link, so I assume I have to put the $search within this if condition?

What would be the best method to pass this info from the options suggested? I'm not too keen on using cookies, but what are the limitations on session variables? Can user browser settings prevent them from working?

yelvington

To represent multiple variables in a URL, separate them with the ampersand character:

foo.php?a=1&b=2&c=whatever

You then can refer to

$GET['a']
$GET['b']
$_GET['c']

To prepare a string for being passed in a URL, you should use
http://us3.php.net/urlencode

nllewellyn

Thank you

I'll try it out 🙂

nllewellyn

Good news..... It worked like a charm 🆒

Thanks for your help